So i have to display out the row in a table that fetch from my database. I have a JQuery to fetch out code but it show me the error Parse error: syntax error, unexpected ';' in C:xampphtdocshopeplacemadminaction.php on line 49 <?php //action.php if(isset($_POST["action"])) { include('../connect.php'); if($_POST["action"] == 'fetch') { $output=''; $query ="SELECT * FROM
所以我必须展示出从我的数据库中获取的表中的行。 我有一个JQuery来获取代码，但它显示我的错误 解析错误：语法错误，意外';' 在第49行的C： xampp htdocs hopeplace madmin action.php中 <?php //action.php if(isset($_POST["action"])) { include('../connect.php'); if($_POST["action"] == 'fetch') { $output=''; $query ="SELECT * FROM user_details WHERE user_type ='user' ORDER BY us
I am trying to access sqlite from PHP script. Not sure how to fetch a single value from table. <?php if ($dbhandle = sqlite_open('c:/sqlite/test.db', 0666, $sqliteerror)) { $sql = "SELECT max (user_id) from persons"; $res = sqlite_query($dbhandle, $sql); if (sqlite_num_rows($result)>0) { echo sqlite_fetch_single($result) ; } } ?> I am getting following er
我试图从PHP脚本访问sqlite。 不确定如何从表中获取单个值。 <?php if ($dbhandle = sqlite_open('c:/sqlite/test.db', 0666, $sqliteerror)) { $sql = "SELECT max (user_id) from persons"; $res = sqlite_query($dbhandle, $sql); if (sqlite_num_rows($result)>0) { echo sqlite_fetch_single($result) ; } } ?> 我得到以下错误 致命错误：调用第3行C： xampp htdocs fetch_d
I have searched the net and so far what I have seen is that you can use mysql_ and mysqli_ together meaning: <?php $con=mysqli_connect("localhost", "root" ,"" ,"mysql"); if( mysqli_connect_errno( $con ) ) { echo "failed to connect"; }else{ echo "connected"; } mysql_close($con); echo "Done"; ?> or <?php $con=mysql_connect("localhost", "root" ,"" ,"mysql"); if( mysqli_connect_e
我搜索了网络，到目前为止我所看到的是，你可以同时使用mysql_和mysqli_这意味着： <?php $con=mysqli_connect("localhost", "root" ,"" ,"mysql"); if( mysqli_connect_errno( $con ) ) { echo "failed to connect"; }else{ echo "connected"; } mysql_close($con); echo "Done"; ?> 要么 <?php $con=mysql_connect("localhost", "root" ,"" ,"mysql"); if( mysqli_connect_errno( $con ) ) { echo "fa
I've looked at other questions about the same error but I'm having trouble applying them to my situation. This is the error I'm getting: Fatal error: Uncaught exception 'Exception' with message 'DateTime::__construct() [datetime.--construct]: Failed to parse time string (2013-07-22164:50:00) at position 10 (1): Unexpected character' in /Applications/XAMPP/xamppfile
我查看了其他有关同样错误的问题，但我无法将它们应用于我的情况。 这是我得到的错误： 致命错误：带有消息'DateTime :: __ construct（）[日期时间.--构造]未捕获异常'异常'：无法解析位置10（1）处的时间字符串（2013-07-22164：50：00）：意外字符'in /Applications/XAMPP/xamppfiles/htdocs/Festival_Planner/index.php:88堆栈跟踪：＃0 /Applications/XAMPP/xamppfiles/htdocs/Festival_Planner/index.
I was running the following PHP code: <?php </script> ?> There were no parse errors and the output was " ?> " (example). In similar cases I do get a parse error: <?php </div> ?> Parse error: syntax error, unexpected '<' in ... Why doesn't <?php </script> ?> give the same error? This must be because there are vario
我正在运行以下PHP代码： <?php </script> ?> 没有解析错误， 输出是 “ ?> ”（例子）。 在类似的情况下，我得到一个解析错误： <?php </div> ?> 解析错误：语法错误，意外的'<'in ... 为什么不<?php </script> ?>给出相同的错误？ 这一定是因为有各种方式启动一段PHP代码： <? ... ?> <? ... ?> （称为short_open_tag ） <?php ... ?
I get this error in my PHP code: PHP Parse error: syntax error, unexpected T_STRING, expecting T_FUNCTION in C:Inetpubwwwrootwebrootwww.novotempo.org.brlibTwitter.php on line 54 The line in question: define('DEBUG',false); Searching the net I found that this usually occurs when you´re using PHP 4.xx, but I´m using 5.2.6 (Just checked it using phpinfo() ). I tried locally, and in two other
我在我的PHP代码中遇到这个错误： PHP解析错误：语法错误，意外的T_STRING，期望在第54行的C： Inetpub wwwroot webroot www.novotempo.org.br lib Twitter.php中的T_FUNCTION 有问题的线路： define('DEBUG',false); 搜索网络时，我发现这通常发生在使用PHP 4.xx时，但我使用的是5.2.6（只是使用phpinfo()检查它）。 我在本地尝试，并在另外两台外部主机上尝试，但它不断返回相同的消息。 为什么会发生？ 我
Possible Duplicate: Warning: mysql_fetch_* expects parameter 1 to be resource, boolean given error I recently got a work from a client, where I have to change the depreciated php old code to new one. In that code I came across mysql_db_query which I converted into mysql_query but the error was given mysql_fetch_array expects parameter 1 to be resource, boolean given //the old code was lik
可能重复： 警告：mysql_fetch_ *期望参数1是资源，布尔给定错误 我最近从客户那里得到了一份工作，我必须将折旧的php旧代码更改为新的代码。 在那段代码中，我遇到了mysql_db_query，我将其转换为mysql_query，但发生错误 mysql_fetch_array期望参数1是资源，布尔给定 //the old code was like: $result = mysql_db_query($mysql_db,"SELECT Hierarchy FROM MenuSystem WHERE LENGTH(Hierarchy) >= 2 AND LOCATE(" .
Possible Duplicate: mysql_fetch_array() expects parameter 1 to be resource, boolean given in select Warning: mysql_fetch_* expects parameter 1 to be resource, boolean given error I was trying to select data from a table but get this error message!! mysql_fetch_array() expects parameter 1 to be resource, boolean given in select I was actually using pagination in php ... but it worked fin
可能重复： mysql_fetch_array（）期望参数1是资源，布尔在select中给出 警告：mysql_fetch_ *期望参数1是资源，布尔给定错误 我试图从表中选择数据，但得到这个错误信息！ mysql_fetch_array（）期望参数1是资源，布尔在select中给出 我实际上在php中使用分页...但它对于更简单的查询很好，只有WHERE中只有一个条件...但是这里...有多个条件，它不起作用完整的代码是： <?php /* Place code to connec
Possible Duplicate: Warning: mysql_fetch_* expects parameter 1 to be resource, boolean given error $result3=mysql_query("select * from $mail ORDER BY `data_id` DESC"); in this case if data_id is NULL query is showing Warning - Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in F:xampphtdocs... If mysql_query fails, it returns false . You need to check that
可能重复： 警告：mysql_fetch_ *期望参数1是资源，布尔给定错误 $result3=mysql_query("select * from $mail ORDER BY `data_id` DESC"); 在这种情况下，如果data_id为NULL，则显示查询警告 - 警告：mysql_fetch_array（）期望参数1是资源，布尔值在F： xampp htdocs中给出... 如果mysql_query失败，则返回false 。 在将$result3传递给任何其他mysql函数之前，您需要检查它。 （在这种情况下也打印或记录错误，以便
This question already has an answer here: Reference - What does this error mean in PHP? 30 answers The problem is your echo on line 7. Your doing <?php echo ... ?> inside of an echo. You should instead concatenate the values into the string. Instead of: echo '<td><form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"><input type="hidden" name="id" va
这个问题在这里已经有了答案： 参考 - 这个错误在PHP中意味着什么？ 30个答案 问题是你在第7行的回声。你在<?php echo ... ?>做<?php echo ... ?> 。 您应该将值连接到字符串中。 代替： echo '<td><form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"><input type="hidden" name="id" value="<?php echo $id; ?>" /><input type="checkbox" name="checki